Why Root Limits Need a Domain Check
When an expression contains a square root or another root, it is tempting to take the limit of the radicand and then take the root of that answer. This works under precise conditions. The conditions matter because even roots of negative real numbers are not real, while odd roots are defined for negative as well as positive real numbers.
The central question is therefore not only what the radicand approaches, but also whether the root expression is defined for real inputs near the point in question. Once that domain check is made, root limits follow from the order properties of powers and roots.
Real Roots and Their Domains
For an even positive integer \(n\), the real \(n\)th root of a number \(u\geq 0\), written \(\sqrt[n]{u}\), is the unique nonnegative real number whose \(n\)th power is \(u\). In particular, \(\sqrt{u}\) means \(\sqrt[2]{u}\). The real domain of an even root is \([0,\infty)\).
For an odd positive integer \(n\), \(\sqrt[n]{u}\) is the unique real number whose \(n\)th power is \(u\). It is defined for every real \(u\), including negative values. For example, \(\sqrt[3]{-8}=-2\), since \((-2)^3=-8\).
The absolute value in the even-root identity is important: an even power removes the sign of its input, and the even root is defined to be nonnegative. In a limit problem, the root must be defined at the nearby input values being considered. For an even root this means its radicand must be nonnegative there.
The Root Limit Rule
Suppose \(g(x)\) approaches a finite real number \(L\) as \(x\) approaches \(a\). The root rule states that the limit of the root is the root of \(L\), provided the real root is defined for \(g(x)\) near \(a\). For even roots, the limiting value \(L\) must also be nonnegative.
Here and below, a limit is taken through values of \(x\) in the domain of the expression. A two-sided limit requires domain points arbitrarily close to \(a\) on both sides. The theorem does not make an expression real-valued at inputs where an even-root radicand is negative.
For even \(n\), the condition \(L\geq 0\) follows if the radicand is nonnegative near \(a\). Indeed, if \(L<0\), then choosing a positive tolerance smaller than \(-L\) would force \(g(x)\) to be negative sufficiently close to \(a\), contradicting \(g(x)\geq 0\). Stating the condition explicitly is useful because it signals whether the proposed real limit makes sense.
Why the Rule Works
The proof uses the fact that taking an \(n\)th root preserves order on its real domain. For even \(n\), the function \(t\mapsto t^n\) is strictly increasing for \(t\geq 0\), so its inverse, the nonnegative \(n\)th-root function, is also strictly increasing. For odd \(n\), the power function is strictly increasing on all real numbers: it is increasing on each side of zero, and a negative number has a negative odd power while a positive number has a positive odd power. Its inverse, the real odd-root function, is therefore strictly increasing as well.
We now prove the limit rule using bounds around the proposed root. This also handles a limiting radicand of zero without dividing by a quantity that could vanish.
Set \(c=\sqrt[n]{L}\), using the real root appropriate to the parity of \(n\). Thus \(c^n=L\).
For odd \(n\), let \(\varepsilon>0\). Strict increase gives \((c-\varepsilon)^n<L<(c+\varepsilon)^n\). For even \(n\), if \(L=0\), use the separate argument below. If \(L>0\), choose \(0<\eta\leq\varepsilon\) with \(\eta<c\); then \(0\leq(c-\eta)^n<L<(c+\eta)^n\).
The two strict gaps between \(L\) and the endpoint powers are positive. Choose \(\delta>0\) smaller than both gaps. If \(0<|x-a|<\delta\), the assumption \(g(x)\to L\) gives the required bound on \(|g(x)-L|\), so \(g(x)\) lies strictly between those endpoint powers.
Since the root function is increasing, the preceding bounds imply \(c-\varepsilon<\sqrt[n]{g(x)}<c+\varepsilon\) in the odd case. In the even case they imply \(c-\eta<\sqrt[n]{g(x)}<c+\eta\), and \(\eta\leq\varepsilon\).
For completeness, in Step 3 choose \(\delta\) so that \(|g(x)-L|<\delta\) implies that \(g(x)\) is between the endpoint powers. For even \(n\) and \(L=0\), given \(\varepsilon>0\), the limit \(g(x)\to 0\) and the domain condition give \(0\leq g(x)<\varepsilon^n\) sufficiently near \(a\). Taking nonnegative \(n\)th roots yields \(0\leq\sqrt[n]{g(x)}<\varepsilon\). In all cases, the root differs from \(c\) by less than the arbitrary tolerance \(\varepsilon\). This is exactly the definition of the stated limit.
Worked Examples
Worked Example: A Square Root with a Positive Limiting Radicand
Evaluate \(\lim_{x\to 2}\sqrt{5x+1}\). First use the constant-multiple and sum laws to find the radicand limit:
The limiting radicand is positive. Also, \(5x+1>0\) whenever \(x\) is sufficiently close to \(2\) (indeed, it is positive for \(x>-1/5\)), so the square root is real near \(2\). Apply the root limit rule:
Check: the proposed answer is nonnegative and \((\sqrt{11})^2=11\), the limiting radicand. Thus the result is consistent with the defining identity for the square root.
Worked Example: An Odd Root with a Negative Limiting Radicand
Evaluate \(\lim_{x\to 3}\sqrt[3]{2x-7}\). By the constant-multiple and sum laws,
The radicand approaches \(-1\). This causes no domain problem because a real cube root is defined for every real radicand. Therefore,
Check: \((-1)^3=-1\), which is the limiting radicand. A negative answer is appropriate for an odd root of a negative number.
Worked Example: An Even Root When the Radicand Tends to Zero
Evaluate \(\lim_{x\to 0}\sqrt[4]{x^2}\). The radicand has limit
For every real \(x\), \(x^2\geq 0\), so the fourth root is defined on both sides of \(0\). The even-root rule at a zero limiting radicand gives
Check: \(0\) is nonnegative and \(0^4=0\). In fact, \(\sqrt[4]{x^2}=|x|^{1/2}\), which also approaches \(0\) as \(x\to 0\).
A Common Domain Mistake
A common error is to apply the even-root rule even when the expression has no real values on one side of the point. For example, the claim
is not a real two-sided limit. Although the radicand tends to \(0\), \(\sqrt{x}\) is not real-valued for \(x<0\), so there are no domain points approaching \(0\) from the left. The valid one-sided statement is
For this right-hand limit, \(x\geq 0\) and \(x\to 0\), so the even-root rule applies. The correction is not to change the computed root; it is to state the limit from the side on which the expression is defined.
Check Your Understanding
For each question, identify the root’s real domain as needed before evaluating or deciding whether the stated two-sided limit exists as a real limit.
- Evaluate \(\lim_{x\to 1}\sqrt[4]{3x+13}\), and state why the root is defined near \(x=1\).
- Evaluate \(\lim_{x\to 2}\sqrt[3]{x-5}\). Why does the negative radicand limit cause no domain problem?
- Evaluate \(\lim_{x\to 0}\sqrt[6]{x^2}\). What domain fact permits a two-sided limit?
- Does \(\lim_{x\to 0}\sqrt{x-1}\) exist as a real two-sided limit? Explain using the radicand’s domain.
- State the condition on \(L\) required for \(\sqrt[n]{L}\) to be a real number when \(n\) is even, and contrast it with odd \(n\).
